532. K-diff Pairs in an Array

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:
Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.
Example 2:
Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
Example 3:
Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
Note:
The pairs (i, j) and (j, i) count as the same pair.
The length of the array won't exceed 10,000.
All the integers in the given input belong to the range: [-1e7, 1e7].

這道題首先二話不說用O(

)的解法秒掉，使用set：

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        const int size = nums.size();
        if(size == 0 || k < 0)
            return 0;

        set<pair<int, int>> s;
        int a[2] = {0};
        for(int i=0; i<size-1; ++i){
            for(int j=i+1; j<size; ++j){
                if(nums[i] - nums[j] == k || nums[j] - nums[i] == k){
                    a[0] = nums[i];
                    a[1] = nums[j];
                    if(a[0] > a[1])
                        std::swap(a[0], a[1]);
                    s.insert(make_pair(a[0], a[1]));
                }
            }
        }
        return s.size();
    }
};

不過有更好的O(n)的解法，使用unordered_set，不過就是要注意k=0的情況要單獨寫出來，只有同一個數出現多次才被計算入結果。